53 109
Assignments Done
97,7%
Successfully Done
In October 2017
Your physics homework can be a real challenge, and the due date can be really close — feel free to use our assistance and get the desired result.
Be sure that math assignments completed by our experts will be error-free and done according to your instructions specified in the submitted order form.
Our experts will gladly share their knowledge and help you with programming homework. Keep up with the world’s newest programming trends.

Answer on Calculus Question for shane duque

Question #4189
A man 6 ft. tall walks away from a lamppost 16 ft. high at a rate of 5 mi per hr. How fast does the of his shadow move?
Expert's answer
Suppose a man was in point A and the shadow was in point S1 at time t = t0. After some time t, the man came to the point B and the shadow – to the point S2. OL is a lamppost.
As we want to evaluate the speed of the shadow, we need to find:
(S1 S0)/t = (S1 B+BS2)/t
https://www.assignmentexpert.com//assignments/uploaded_files/static/4b/08/4b0817ee471397f033b5e77965b53c88.bmp

Angles <OS1L and <CS1A are equal, so:
OL/(LA+AS1 )=AC/(AS1 )
AS1*H = AS1*h + LA*h
Where H is the height of the lamppost (H = 16ft) and h is the man’s height (h = 6ft).
AS1 = LA*h/(H-h)
So, S1 B = AB - AS1 = Vt - LA*h/(H-h)
Angles <OS2 L and <DS2 B are equal, so:
OL/(LA+AB+BS2 ) = BD/(BS2 )
BS2*H = LA*h + Vt*h + BS2*h
BS2 = h(LA+Vt)/(H-h)
Then:
(S1 S2)/t = (S1 B+BS2)/t = V - LA/t*h/(H - h) + h(LA + Vt)/(H - h)t = V + (h*LA + Vth - LA*h)/((H - h)t)
= V + V*h/(H - h) = VH/(H - h)
The speed of the shadow is 5*16/(16-6)=8 miles/hour

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question